Algorithm solves a problem of size 𝑛 as follows: recursively solves 3 subproblems of size 𝑛 - 2, and then constructs the answer for the original problem in time 𝑂(1).
It seems obvious that the Master theorem cannot be applied here. So, I thought of drawing a recursion tree, but what bothers me is that do I need to consider two cases: when n is odd/even? Or the result sum and hence the running time wouldn't depend on this at all? Thanks in advance.
We'll have to assume that the base case of the recursion occurs when 𝑛 is either 0 or 1. It could also be that the base case occurs when 𝑛 is either 1 or 2, 𝑛 is not allowed to be 0. We don't really know, but it is not that relevant.
In the first case, the number of operations for a given 𝑛 that is odd, is the same as the number of operations for 𝑛−1. In the second case it would be the same number of operations as for 𝑛+1.
So, for determining the asymptotic complexity we can look at just the even numbers (or only the odd numbers).
The recursion tree is a perfect 3-ary tree with a height of (𝑛+1)/2 or (𝑛+2)/2 (again: depending on the base case).
In a perfect 3-ary tree of height h, we have (3h-1)/2 nodes, so that is O(3h), which in terms of 𝑛 is O(3𝑛/2) = O((√3)𝑛).
Another (possibly simpler) solution in addition to the one proposed by @trincot :
T(n)=3T(n-2)+O(1)=...=3k(T(n-2k)+O(1)).
Let's find out the stopping condition for this recurrence relation, i.e. need to find k for which the following holds: n-2k=0 --> k=n/2 --> T(n)=O(3n/2)=O(√3n)
